robustness of fuzzy reasoning via logically equivalence measure

Information Sciences 177 (2007) 5103–5117

www.elsevier.com/locate/ins

Robustness of fuzzy reasoning via logicallyequivalence measure q

Jianhua Jin a,c, Yongming Li a,b,*, Chunquan Li a,c

a College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, Chinab College of Computer Science, Shaanxi Normal University, Xi’an 710062, Chinac College of Sciences of Southwest Petroleum University, Chengdu 610000, China

Received 1 October 2006; received in revised form 1 May 2007; accepted 2 May 2007

Abstract

In this paper, we discuss robustness of fuzzy reasoning. After proposing the definition of perturbation of fuzzy setsbased on some logic-oriented equivalence measure, we present robustness results for various fuzzy logic connectives, fuzzyimplication operators, inference rules and fuzzy reasoning machines, and discuss the relations between the robustness offuzzy reasoning and that of fuzzy conjunction and implication operators. The robustness results are presented in terms ofd-equalities of fuzzy sets based on some logic-oriented equivalence measure, and the maximum of d (which ensures thecorresponding d-equality holds) is derived.� 2007 Elsevier Inc. All rights reserved.

Keywords: Fuzzy sets; Fuzzy reasoning; Robustness of fuzzy reasoning; Logically equivalence; d-equality

1. Introduction

After Zadeh introduced fuzzy sets [12], fuzzy reasoning or approximate reasoning [5,13,14] has been anactive research pursuit, which is extensively used in intelligent systems including fuzzy control, classification,expert systems, and networks to name a few dominant categories of such architectures. Various methods offuzzy reasoning (see [4,6,9,10]) have been proposed along with numerous alternative realizations of implica-tion operators and logic connectives (see [2,3]). Among those methods of fuzzy reasoning, the results of fuzzyreasoning are dependent on the choice of fuzzy sets of fuzzy antecedent and fuzzy consequences as well asfuzzy connectives and fuzzy implication operators which link fuzzy antecedents and fuzzy consequences. In

0020-0255/$ - see front matter � 2007 Elsevier Inc. All rights reserved.

doi:10.1016/j.ins.2007.05.014

q This work is supported by National Science Foundation of China (Grant No. 10571112), National 973 Foundation Research Program(Grant No. 2002CB312200) and Key Research Project of Ministry of Education of China (Grant No. 107106).

* Corresponding author. Address: College of Computer Science, Shaanxi Normal University, Xi’an 710062, China. Tel.: +86 2985307628; fax: +86 29 85310161.

E-mail addresses: [email protected] (J. Jin), [email protected] (Y. Li).

mailto:[email protected]

mailto:[email protected]

5104 J. Jin et al. / Information Sciences 177 (2007) 5103–5117

the research area of fuzzy control, one of the most important problems is the analysis of stability and robust-ness of fuzzy controllers. From the viewpoint of fuzzy reasoning, there has a corresponding problem: will asmall variance of input cause a big variance of output of fuzzy reasoning?

To address this problem, Ying [11] proposed the concepts of maximum and average robustness of fuzzysets. Subsequently perturbation parameters of various methods of fuzzy reasoning could be estimated. Thoseperturbation parameters could help formulate certain criteria effectively in selecting a specific method of fuzzyreasoning. Cai [1] used the notion of d-equalities of fuzzy sets to study robustness of fuzzy reasoning that ledto some general results in case of fuzzy connectives, fuzzy implication operators, generalized modus ponensand generalized modus tollens. Noting that the robustness of fuzzy reasoning significantly relies on the choiceof the fuzzy connectives, Li et al. [7,8] presented an approach to measure the robustness of fuzzy reasoningusing the sensitivity of fuzzy connectives, and derived the best perturbation parameters of fuzzy reasoningthere.

In the previous work, the perturbation of fuzzy sets was expressed based on the notion of the maximumperturbation or d equalities of fuzzy sets via the distance measure on the unit interval [0, 1]. However, thebehavior of a fuzzy logic system is mainly determined by its internal logic structure—the fuzzy connectivesand fuzzy implication operators. Having this in mind, we propose another form of perturbation based on log-ically equivalence measure or similarity of fuzzy sets A and B (see Definition 1.1 below) [8]:

A � B ¼ ðA! BÞHðB! AÞ;

where ! is an implication operator, and w is a t-norm.

Definition 1.1. Let X be a nonempty set. If A;B 2 F ðX Þ, 0 6 d 6 1 and ! is a fuzzy connective, then we saythat A;B are d-equal based on logically equivalence measure, which denoted by A � ðdÞB in symbols, if thefollowing condition holds: ½A � B� ¼

Vx2X ðAðxÞ ! BðxÞÞ ^ ðBðxÞ ! AðxÞÞP d. In this case, B is called a 1� d

perturbation of A.

Lemma 1.1. For any x; y 2 ½0; 1�, let! be Lukasiewicz implication operator, that is, x! y ¼ minf1; 1� xþ yg.Then ½x � y� ¼ 1� jx� yj.

Lemma 1.2. Let X be a nonempty set. If A;B 2 F ðX Þ, 0 6 d 6 1 and! is Lukasiewicz implication operator, then

the following conditions are equivalent:

(1)W

x2X jAðxÞ � BðxÞj 6 1� d, i.e., B is a 1� d maximum perturbation of A as defined in [8].(2) A � ðdÞB.

From Lemma 1.2, it shows that perturbation of fuzzy sets based on logically equivalence measure is morecomprehensive than that of fuzzy sets based on the notion of the maximum perturbation via the distance mea-sure on the unit interval [0, 1]. That is, the perturbation of fuzzy sets which discussed in [1,8,11] is a special caseof Definition 1.1 which based on logically equivalence measure when we choose the fuzzy connective ! asLukasiewicz implication operator. Therefore, we will get more general results about the robustness of fuzzyreasoning systems in this paper. Of course, to obtain some nice results about the robustness of fuzzy reasoningsystems based on logically equivalence measure introduced in this paper, we have to give some restrictions onthe choices of fuzzy logical connectives, which include the often used fuzzy logical connectives in the practicalapplications.

The remainder part is organized in the following way. In Section 2, the robustness of fuzzy modus pon-ens is studied, and some general results concerning robustness of some logic connectives are investigated. InSection 3, we study the robustness of fuzzy modus tollens. In Section 4, we study the robustness of the keycategories of fuzzy inference systems. An illustrative example is presented in Section 5. Conclusions are pre-sented in Section 6. Throughout the study we use a standard notation encountered in fuzzy sets. In par-ticular we use sup(_) to represent the supremum operation while inf(^) is used to represent the infimumoperation.

J. Jin et al. / Information Sciences 177 (2007) 5103–5117 5105

2. Robustness of fuzzy modus ponens

Before we proceed with the development of the perturbation parameters of fuzzy reasoning, we recallZadeh’s compositional rule of inference. The general form of multidimensional fuzzy reasoning may be statedas follows:

Premise 1: If x1 is A1 and x2 is A2 and � � � and xn is An, then y is B;Premise 2: x1 is A01 and x2 is A02 and � � � and xn is A0n;Conclusion: y is B 0.

where X 1; . . . ;X n, and Y are n + 1 universes of discourse, Ai and A0i are fuzzy subsets of X iði ¼ 1; . . . ; nÞ, and B

and B 0 are fuzzy subsets of Y. By R we denote a fuzzy relation between X and Y, A and A 0 as fuzzy relationsamong Xi ði ¼ 1; . . . ; nÞ, i.e. , R ¼ A! B, A ¼ A1 � A2 � � � � � An, where AðxÞ ¼ A1ðx1Þ � A2ðx2Þ � � � � � AnðxnÞ,and A0ðxÞ ¼ A01ðx1Þ � A02ðx2Þ � � � � � A0nðxnÞ for x ¼ ðx1; x2; . . . ; xnÞ 2 X . In [12], Zadeh proposed the so-called com-positional rule of inference to implement the above fuzzy reasoning in the following way: B0 ¼ A0 � ðA! BÞ,where B0ðyÞ ¼

Wx2X ½A0ðxÞ � ðAðxÞ ! BðyÞÞ� for any y 2 Y , and * stands for a t-norm, A! B for a fuzzy relation

from X to Y defined by imposing an implication operator on A and B. In this section we discuss robustness offuzzy modus ponens under various specific t-conjunctions and implication operators.

Definition 2.1. For any A1i;A2i;A01i;A02i 2 F ðX iÞ, B1;B2;B01;B

02 2 F ðY Þði ¼ 1; . . . ; nÞ, let Aj ¼

Qni¼1Aji,

A0j ¼Qn

i¼1A0ji, Rjðx; yÞ ¼ AjðxÞ ! BjðyÞ, B0j ¼ A0j � Rjðj ¼ 1; 2Þ and 0 6 d 6 1. The perturbation parameters4B0 ðdÞ of fuzzy reasoning B0 ¼ A0 � R is defined as follows:

4B0 ðdÞ ¼ inff½B01 � B02�jA1i � ðdÞA2i;A01i � ðdÞA02i;B1 � ðdÞB2; i ¼ 1; . . . ; ng:

Definition 2.2. Let f : ½0; 1�m ! ½0; 1� be a fuzzy operator. For any d 2 ½0; 1�, if there exist x; y 2 ½0; 1�m,½xi � yi�P dð1 6 i 6 mÞ such that Mf ðdÞ ¼ ½f ðxÞ � f ðyÞ�, then we call f an perturbation attainable operator.

Concerning the robustness of fuzzy reasoning, we mainly focus on perturbable attainable operators. Usu-ally, the t-norm * will be minimum, product or Lukasiewicz conjunction (i. e., x � y ¼ maxf0; xþ y � 1g), andthe implication ! will be Lukasiewicz, Goguen, Mamdani or Kleene-Dienes implication, that is,

x! y ¼ minf1; 1� xþ yg; x!GOy ¼ 1; if x ¼ 0;yx ^ 1; if x 6¼ 0;

�x! y ¼ x ^ y or x! y ¼ ð1� xÞ _ y, x; y 2 ½0; 1�.

Clearly, they are attainable. Specially, according to Definition 2.1, we do not restrain to the structure of fuzzyset. So the following result is obvious.

Lemma 2.1. If f ðx01; . . . ; x0n; x1; . . . ; xn; yÞ ¼ ðx01�0 � � � �0x0nÞ � ðx1�0 � � � �0xn ! yÞ ¼ x0 � ðx! yÞ, and f is a pertur-bation attainable fuzzy operator, then 4f ðdÞ ¼ 4B0 ðdÞ.

Given this lemma, it is sufficient to substitute the research of perturbation of the corresponding fuzzy oper-ator for that of fuzzy reasoning systems. Firstly we discuss case of n = 1.

Lemma 2.2. For any x; y 2 ½0; 1�, the following statements hold:

(1) Suppose x! y ¼ x ^ y. Then ½x � y�P d if and only if x P d and y P d:(2) Suppose x! y ¼ ð1� xÞ _ y. Then ½x � y�P d if and only if either 1� x P y, 1� x P d and 1� y P d or

1� x 6 y, x P d and y P d.

(3) Consider the Goguen implication. Then ½x � y�P d if and only if either yd 6 x 6 y or xd 6 y 6 x. Spe-

cially, for d 2 ð0; 1�, ½x � y�P d if and only if dx 6 y 6 xd ^ 1.

Remark 2.1. If ½A � B�P d, then by Lemmas 2.1 and 2.2, the following statements are valid:

(i) When ! is chosen as Lukasiewicz or Goguen implication, the bigger dðd 2 ½0; 1�Þ is, the closer the fuzzysets A and B are.

(ii) When! is chosen as Mamdani or Kleene-Dienes implication, the bigger dðd 2 ½0; 1�Þ is, the closer and thecrisper the fuzzy sets A and B are.


Theorem 2.1. Let f ðx; y; zÞ ¼ x � ðy ! zÞ. Then the following statements hold:

(1) Suppose * is the min conjunction. Then, (a) if! is the Lukasiewicz implication, then4f ðdÞ ¼ ð2d� 1Þ _ 0;

(b) if! is the Mamdani or Kleene-Dienes implication, then 4f ðdÞ ¼ d; (c) if! is the Goguen implication,

then 4f ðdÞ ¼ d2.

(2) Suppose * is the product conjunction. Then, (a) if ! is the Lukasiewicz implication, then

4f ðdÞ ¼ ð2d2 � dÞ _ 0; (b) if ! is the Mamdani or Kleene-Dienes implication, then 4f ðdÞ ¼ d2; (c) if !is the Goguen implication, then 4f ðdÞ ¼ d3.

(3) Suppose * is the Lukasiewicz conjunction. Then, (a) if ! is the Lukasiewicz implication, then4f ðdÞ ¼ ð3d� 2Þ _ 0; (b) if ! is the Mamdani or Kleene-Dienes implication, then 4f ðdÞ ¼ ð2d� 1Þ _ 0.

Proof

(1)(a) Suppose x � y ¼ minðx; yÞ, x! y ¼ minð1; 1� xþ yÞ, x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0. By Lemma 1.1and Lemma 1.2, we conclude that ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ 1� jf ðx; y; zÞ � f ðx0; y0; z0Þj ¼ 1� jx^1 ^ ð1� y þ zÞ � x0 ^ 1 ^ ð1� y 0 þ z0ÞjP ð1� jx� x0j _ j1� 1j _ jy0 � y þ z� z0jÞ _ 0 P ð1� jx� x0j_ðjy0 � yj þ jz� z0jÞÞ _ 0 P ð1� ð1� dÞ _ ð1� dþ 1� dÞÞ _ 0 ¼ ð2d� 1Þ _ 0. In particular, whenx = 1, x0 ¼ d, y ¼ d, y0 ¼ 1, z ¼ 1� d, z0 ¼ 0, all of the above equalities hold. So4f ðdÞ ¼ ð2d� 1Þ _ 0.

(2)(a) Suppose x � y ¼ xy, and x! y ¼ minð1; 1� xþ yÞ, x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0. By the symmetryof f ðx; y; zÞ and f ðx0; y0; z0Þ, without loss of generality we assume that f ðx; y; zÞP f ðx0; y0; z0Þ. Then byLemma 1.1 and Lemma 1.2, we know that ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ 1� jf ðx; y; zÞ � f ðx0; y0; z0Þj ¼1� f ðx; y; zÞ þ f ðx0; y 0; z0Þ ¼ 1� xðy ! zÞ þ x0ðy 0 ! z0ÞP 1� xðy ! zÞ þ ðx� ð1� dÞÞðy ! z� kÞ ¼1� xk� ð1� dÞðy ! z� kÞP 1� k� ð1� dÞð1� kÞ ¼ d� dk, where k ¼

Wjy ! z� y0 ! z0j ¼ 2ð1�

dÞ ^ 1. So ½f ðx; y; zÞ � f ðx0; y0; z0Þ�P 0 _ ð2d2 � dÞ. Moreover, when x = 1, x0 ¼ d, y ¼ z ¼ d, y 0 ¼ 1,z0 ¼ ð2d� 1Þ _ 0, the equality holds. So 4f ðdÞ ¼ ð2d2 � dÞ _ 0.

(3)(a) Suppose x � y ¼ 0 _ ðxþ y � 1Þ, x! y ¼ minð1; 1� xþ yÞ, x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0. Then byLemma 1.1 and Lemma 1.2, we conclude that ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ 1� jf ðx; y; zÞ � f ðx0;y0; z0Þj ¼ 1� j0 _ ðxþ y ! z� 1Þ � 0 _ ðx0 þ y 0 ! z0 � 1ÞjP 1� jðxþ y ! z� 1Þ � ðx0 þ y0 ! z0 � 1ÞjP 0 _ ð1� ½jy ! z� y0 ! z0j þ jx� x0j�ÞP 0 _ ð1� 3ð1� dÞÞ ¼ ð3d� 2Þ _ 0. Moreover, when x = 1,x0 ¼ d, y ¼ z ¼ d, y0 ¼ 1, z0 ¼ ð2d� 1Þ _ 0, all of the above equalities hold. So 4f ðdÞ ¼ ð3d� 2Þ _ 0.

(1)(b) If x � y ¼ x ^ y, x! y ¼ x ^ y, x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0, then it follows from Lemma 2.2(1)that x P d, x0 P d, y P d, y0 P d, z P d and z0 P d. And thus ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ x^y ^ z ^ x0 ^ y 0 ^ z0 P d. Particularly, when x ¼ y ¼ z ¼ x0 ¼ y0 ¼ z0 ¼ d, the equality holds. So4f ðdÞ ¼ d.If x � y ¼ x ^ y, x! y ¼ ð1� xÞ _ y, then ½f ðx; y; zÞ � f ðx0; y 0; z0Þ� ¼ ½ð1� x ^ ðð1� yÞ _ zÞÞ _ ðx0^ðð1� y0Þ _ z0ÞÞ� ^ ½ð1� x0 ^ ðð1� y0Þ _ z0ÞÞ _ ðx ^ ðð1� yÞ _ zÞÞ�, let l ¼ ð1� x ^ ðð1� yÞ _ zÞÞ _ ðx0^ðð1� y0Þ _ z0ÞÞ, m ¼ ½ð1� x0 ^ ðð1� y 0Þ _ z0ÞÞ _ ðx ^ ðð1� yÞ _ zÞÞ�. Since x � ðdÞx0, y � ðdÞy0 andz � ðdÞz0, it follows from Lemma 2.2(2) that ð1� xÞ _ x0 P d, ð1� y0Þ _ y P d and ð1� zÞ _ z0 P d.And thus l ¼ ðð1� xÞ _ ðy ^ ð1� zÞÞÞ _ ðx0 ^ ðð1� y0Þ _ z0ÞÞ ¼ ½ð1� xÞ _ x0 _ ðy ^ ð1� zÞÞ�^½ð1� xÞ _ y _ ð1� y0Þ _ z0� ^ ½ð1� xÞ _ ð1� zÞ _ ð1� y 0Þ _ z0�P d. Using the same method we can con-clude that m P d. Hence ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ l ^ m P d. For d P 1

2, let x ¼ y ¼

z ¼ x0 ¼ y0 ¼ z0 ¼ d, then the equality holds. For d 6 12, let x ¼ y ¼ z ¼ 1� d, x0 ¼ y0 ¼ z0 ¼ d, then

the equality holds. So 4f ðdÞ ¼ d.(2)(b) If x � y ¼ xy, x! y ¼ x ^ y, x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0, then by Lemma 2.2(1), it follows that

x P d, x0 P d, y P d, y0 P d, z P d and z0 P d. And thus ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ½xðy ^ zÞ�^½x0ðy0 ^ z0Þ�P d2. Let x ¼ y ¼ z ¼ x0 ¼ y 0 ¼ z0 ¼ d. Then the equality holds. So 4f ðdÞ ¼ d2.If x � y ¼ xy and x! y ¼ ð1� xÞ _ y, then ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ½ð1� xðð1� yÞ _ zÞÞ _ ðx0ðð1� y0Þ _ z0ÞÞ� ^ ½ð1� x0ðð1� y0Þ _ z0ÞÞ _ ðxð1� yÞ _ zÞÞ�. Since x � ðdÞx0, by Lemma 2.2(2) we con-clude that either 1� x P x0; 1� x P d and 1� x0 P d, or 1� x 6 x0; x P d and x0 P d; ½y � y0�P dis equivalent to ðy; y0Þ 2 A [ B, and ½z � z0�P d is equivalent to ðz; z0Þ 2 C [ D, where


A ¼ fðy; y0Þj1� y P d; 1� y 0 P d; 1� y P y0g, B ¼ fðy; y 0Þjy P d; y0 P d; 1� y 6 y 0g,C ¼ ðz; z0Þj1� z P d; 1� z0 P d; 1� z P z0g, and D ¼ fðz; z0Þjz P d; z0 P d; 1� z 6 z0g.If 1� xðð1� yÞ _ zÞÞP x0ðð1� y0Þ _ z0ÞÞ, then ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ½1� xðð1� yÞ _ zÞÞ�^½1� x0ðð1� y0Þ _ z0Þ�. There are the following cases:Case 1: 1� x P x0; 1� x P d and 1� x0 P d. Then ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ½1� xðð1� yÞ _ zÞÞ�^½1� x0ðð1� y0Þ _ z0Þ�P ð1� xÞ ^ ð1� x0ÞP d. Let x ¼ 1� d, x0 ¼ 0, y ¼ 0, z ¼ 1, y0 ¼ 1� d, z0 ¼ d.Then all of the above equalities hold.Case 2: 1� x 6 x0; x P d and x0 P d. If ðy; y0Þ 2 A; ðz; z0Þ 2 C [ D or ðy; y0Þ 2 B; ðz; z0Þ 2 D, thenð1� yÞ _ z P d and ð1� y0Þ _ z0 P d. It follows that d2

6 x0ðð1� y0Þ _ z0ÞÞ 6 1� xðð1� yÞ _ zÞ 61� d2 and d2

6 xðð1� yÞ _ zÞÞ 6 1� x0ðð1� y0Þ _ z0ÞÞ 6 1� d2. So, for d 2 ½0;ffiffi2p

2�,

½f ðx; y; zÞ � f ðx0; y 0; z0�P d2; for d 2 ðffiffi2p

2; 1�, 1� xðð1� yÞ _ zÞ 6 1� d2 < d2

6 x0ðð1� y 0Þ _ z0ÞÞ, whichcontradicts the assumption. If ðy; y0Þ 2 B and ðz; z0Þ 2 C, then ð1� yÞ _ z 6 1� d andð1� y0Þ _ z0 6 1� d. It follows that 1� xðð1� yÞ _ zÞP 1� xð1� dÞP 1� ð1� dÞ ¼ d, and1� x0ðð1� y0Þ _ z0ÞP 1� x0ð1� dÞP 1� ð1� dÞ ¼ d. Hence ½f ðx; y; zÞ � f ðx0; y 0; z0Þ�P d. If wechoose x ¼ 1; y ¼ d; z ¼ 1� d; x0 ¼ d; y0 ¼ 1; z0 ¼ 0, then the equality holds, and the inequality1� xðð1� yÞ _ zÞÞP x0ðð1� y0Þ _ z0Þ also holds.

On the other hand, if ð1� xðð1� yÞ _ zÞÞ 6 ðx0ðð1� y0Þ _ z0Þ, then 1� x 6 x0. By Lemma 2.2(2), weknow that x P d and x0 P d are valid simultaneously. Now ½f ðx;y; zÞ � f ðx0;y0; z0Þ� ¼ ½x0ðð1� y0Þ _ z0Þ�^½xðð1� yÞ _ zÞ�. There are two cases to be considered here.If ‘‘ðy; y0Þ 2 A; ðz; z0Þ 2 C [ D’’ or ‘‘ðy; y0Þ 2 B; ðz; z0Þ 2 D’’, we could derive that ð1� yÞ _ z P d andð1� y0Þ _ z0 P d. Hence ½f ðx; y; zÞ � f ðx0; y0; z0Þ�P dd ¼ d2. Let x ¼ d, x0 ¼ z0 ¼ 1, z ¼ d, y ¼ 1� d,y0 ¼ 0. Then the equality holds.If ðy; y0Þ 2 B; ðz; z0Þ 2 C, then ð1� yÞ _ z 6 1� d and ð1� y0Þ _ z0 6 1� d, and thus 1� d P x0ðð1�y0Þ _ z0Þ P 1 � xðð1 � yÞ _ zÞÞ P 1 � xð1 � dÞ P 1 � ð1 � dÞ ¼ d; 1 � d P xðð1 � yÞ _ zÞÞ P 1 � x0

ðð1� y0Þ _ z0ÞÞP 1� x0ð1� dÞP 1� ð1� dÞ ¼ d. So for d 2 ½0; 12�, ½f ðx; y; zÞ � f ðx0; y0; z0Þ�P d. In

particular, if we choose x ¼ 1; x0 ¼ 1; y ¼ 1; y0 ¼ d; z ¼ d; z0 ¼ 1� d, then the equality holds; ford 2 ð12 ; 1�, x0ðð1� y0Þ _ z0ÞÞ 6 1� d < d 6 1� xðð1� yÞ _ zÞ, which contradicts the assumption.In summary, since d P d2, we conclude the equality 4f ðdÞ ¼ d2.

(3)(b) If x � y ¼ 0 _ ðxþ y � 1Þ, x! y ¼ x ^ y, x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0, then by Lemma 2.2(1), itfollows that x P d, x0 P d, y P d, y0 P d, z P d and z0 P d. Hence, ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ð0 _ ðxþ y ^ z� 1ÞÞ ^ ð0 _ ðx0 þ y0 ^ z0 � 1ÞÞP 0 _ ð2d� 1Þ. Let x ¼ y ¼ z ¼ x0 ¼ y0 ¼ z0 ¼ d. Then the equal-ity holds. Therefore, 4f ðdÞ ¼ 0 _ ð2d� 1Þ.

If x! y ¼ ð1� xÞ _ y, then ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ½ð1� x � ðy ! zÞÞ _ ðx0 � ðy0 ! z0ÞÞ� ^ ½ð1� x0�ðy0 ! z0ÞÞ _ ðx � ðy ! zÞÞ�. Since x � ðdÞx0, by Lemma 2.2(2) we conclude that either 1� x P x0; 1� x P dand 1� x0 P d, or 1� x 6 x0; x P d and x0 P d. By simple calculation, we also know that ½y � y0�P d isequivalent to ðy; y0Þ 2 A [ B, and ½z � z0�P d is equivalent to ðz; z0Þ 2 C [ D, where A ¼ fðy; y0Þj1� y Pd; 1� y0 P d; 1� y P y0g, B ¼ fðy; y0Þjy P d; y0 P d; 1� y 6 y0g, C ¼ ðz; z0Þj1� z P d; 1� z0 P d; 1� z Pz0g, and D ¼ fðz; z0Þjz P d; z0 P d; 1� z 6 z0g.

If 1� x � ðy ! zÞP x0 � ðy0 ! z0Þ, then ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ð1� x � ðy ! zÞÞ ^ ð1� x0 � ðy0 ! z0ÞÞ ¼ð1 ^ ð2� x� ð1� yÞ _ zÞÞÞ ^ ð1 ^ ð2� x0 � ð1� y0Þ _ z0ÞÞ. Two cases have to be discussed here.

Case 1: 1� x P x0; 1� x P d and 1� x0 P d. In this case, we have ½f ðx; y; zÞ � f ðx0; y0; z0�Pð1� xÞ ^ ð1� x0ÞP d. Let x ¼ 1� d; x0 ¼ y ¼ 0; y0 ¼ 1� d; z ¼ 1; z0 ¼ d. Then all of the above equalitieshold.

Case 2: 1� x 6 x0; x P d and x0 P d. If ‘‘ðy; y0Þ 2 A; ðz; z0Þ 2 C [ D’’ or ‘‘ðy; y0Þ 2 B; ðz; z0Þ 2 D’’, it followsthat ð1� yÞ _ z P d and ð1� y0Þ _ z0 P d. Hence 0 _ ð2d� 1Þ 6 0 _ ðx0 þ ð1� y0Þ _ z0 � 1Þ ¼ x0 � ðy0 !z0Þ 6 1� x � ðy ! zÞ ¼ 1 ^ ð2� x� ð1� yÞ _ zÞ 6 1 ^ ð2� 2dÞ; 0 _ ð2d� 1Þ 6 0 _ ðxþ ð1� yÞ _ z� 1Þ 6 1^ð2� x0 � ð1� y0Þ _ z0 6 1 ^ ð2� 2dÞ. So if d 2 ½0; 3

4�, then ½f ðx; y; zÞ � f ðx0; y0; z0�P 0 _ ð2d� 1Þ; else ifd 2 ð34 ; 1�, then 1� x � ðy ! zÞ ¼ 1 ^ ð2� x� ð1� yÞ _ zÞ 6 1 ^ ð2� 2dÞ < 0 _ ð2d� 1Þ 6 0 _ ðx0 þ ð1� y0Þ_z0 � 1Þ ¼ x0 � ðy0 ! z0Þ, which contradicts the assumption. If ‘‘ðy; y0Þ 2 B; ðz; z0Þ 2 C’’, then ð1� yÞ _ z 6 1� dand ð1� y0Þ _ z0 6 1� d. It follows that 1 ^ ð2� x� ð1� yÞ _ zÞP 1 ^ ð2� 1� ð1� dÞÞ ¼ d. 1 ^ ð2� x0�ð1� y0Þ _ z0ÞP 1 ^ ð2� 1� ð1� dÞÞ ¼ d. Hence ½f ðx; y; zÞ � f ðx0; y0; z0Þ�P d. Let x ¼ 1; y ¼ d; z ¼


1� d; x0 ¼ d; y0 ¼ 1; z0 ¼ 0. Then the above equality holds, and the condition 1 ^ ð2� x� ð1� yÞ _ zÞP0 _ ðx0 þ ð1� y0Þ _ z0 � 1Þ also holds.

On the other hand, if 1� x � ðy ! zÞ 6 x0 � ðy0 ! z0Þ, then 1� x 6 x0. It follows that x P d and x0 P d,½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ðx � ðy ! zÞÞ ^ ðx0 � ðy0 ! z0ÞÞ ¼ 0 _ ½ðx0 þ ð1� y0Þ _ z0 � 1Þ ^ ðxþ ð1� yÞ _ z� 1Þ�.Two cases have to be discussed here.

Case 1: ‘‘ðy; y0Þ 2 A; ðz; z0Þ 2 C [ D’’ or ‘‘ðy; y0Þ 2 B; ðz; z0Þ 2 D’’. It follows that ð1� yÞ _ z P d; ð1� y0Þ_z0 P d. Hence ½f ðx; y; zÞ � f ðx0; y0; z0Þ�P 0 _ ð2d� 1Þ. Let x ¼ d; x0 ¼ 1; z0 ¼ 1; z ¼ d; y ¼ d; y0 ¼ 1. Thenthe above equality holds, and the condition 1� x � ðy ! zÞ 6 x0 � ðy0 ! z0Þ is also satisfied.

Case 2: ‘‘ðy; y0Þ 2 B; ðz; z0Þ 2 C’’. It follows that ð1� yÞ _ z 6 1� d; ð1� y0Þ _ z0 6 1� d, 1� d P x0 � ðy0 !z0Þ ¼ 0 _ ½ðx0 þ ð1� y0Þ _ z0 � 1ÞP 1� x � ðy ! zÞ ¼ 1 ^ ð2� x� ð1� yÞ _ zÞÞP 1 ^ ð2� 1� ð1� dÞÞ ¼ d.1� d P x � ðy ! zÞP 1� x0 � ðy0 ! z0Þ ¼ 1 ^ ð2� x0 � ð1� y0Þ _ z0ÞP 1 ^ ð2� 1� ð1� dÞÞ ¼ d. So ifd 2 ½0; 1

2�, then ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ðx � ðy ! zÞÞ ^ ðx0 � ðy0 ! z0ÞP d. However, if d 2 ð12 ; 1�, thenx0 � ðy0 ! z0Þ ¼ 0 _ ½ðx0 þ ð1� y0Þ _ z0 � 1Þ 6 1� d < d 6 1 ^ ð2� x� ð1� yÞ _ zÞÞ ¼ 1� x � ðy ! zÞ, whichcontradicts the condition.

In summary, it follows that Mf ðdÞ ¼ ð2d� 1Þ _ 0.(1)(c) Suppose d 2 ð0; 1�; x � y ¼ minðx; yÞ, x!GOy ¼ 1; if x ¼ 0;

yx ^ 1; if x 6¼ 0;

�x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0.

Then by Lemma 2.2, we conclude that dt0 6 t 6 1 ^ t0d, dt 6 t0 6 1 ^ t

d ; 8t 2 fx; y; zg.If any element of x; y and z is equal to zero, then the value of ½f ðx; y; zÞ � f ðx0; y0; z0Þ� is either equal to 1 or

not less than d. Thus the following discussion is studied under the condition that all of x; y; z; x0; y0 and z0 arebelong to ð0; 1�. There are two following cases:

Case 1: y 6 z. So f ðx; y; zÞ ¼ x. Then there are two following sub-cases: y0 6 z0, or z0 6 y0. If y0 6 z0, thenf ðx0; y0; z0Þ ¼ x0. Thus ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ½x � x0�P d. If z0 6 y0, then f ðx0; y0; z0Þ ¼ x0 ^ z0

y0. It follows thatf ðx;y;zÞ

f ðx0;y0;z0Þ ¼ xx0^z0

y0¼ x

x0 _xy0

z0 P d, and

f ðx0; y 0; z0Þf ðx; y; zÞ ¼

x0 ^ z0

y0

x¼ x0

x^ z0

xy 0P

x0

x^ z0

xðyd ^ 1Þ ¼x0

x^ ðdz0

xy_ z0

xÞP x0

x^ dz0

zP d ^ d2 ¼ d2:

Thus

½f ðx; y; zÞ � f ðx0; y 0; z0Þ� ¼ 1 ^ f ðx; y; zÞf ðx0; y0; z0Þ ^

f ðx0; y0; z0Þf ðx; y; zÞ P d2:

Case 2: z 6 y. So f ðx; y; zÞ ¼ x ^ zy. There are two sub-cases to be discussed: y 0 6 z0, or z0 6 y 0. If y 0 6 z0, then

f ðx0; y 0; z0Þ ¼ x0. It follows that f ðx;y;zÞf ðx0 ;y0;z0Þ ¼

x^zy

x0 ¼ xx0 ^ z

x0y P d ^ dz0

x0y P d ^ dy0

y P d2, and f ðx0 ;y0;z0Þf ðx;y;zÞ ¼ x0

x^zy¼ x0

x _x0yz P d.

Thus ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ 1 ^ f ðx;y;zÞf ðx0 ;y0;z0Þ ^

f ðx0;y0;z0Þf ðx;y;zÞ P d2. If we choose x ¼ y0 ¼ z0 ¼ d; x0 ¼ y ¼ 1; z ¼ d2,

then ½f ðx; y; zÞ � f ðx0; y 0; z0Þ� ¼ d2: If z0 6 y0, then f ðx0; y 0; z0Þ ¼ x0 ^ z0

y0. It follows that f ðx;y;zÞf ðx0 ;y0 ;z0Þ ¼ x^

zy

x0^z0y0¼ð

xx0_

xy0z0 Þ^ð

zx0y_

zy0z0yÞPd^d2¼d2

, and f ðx0;y0 ;z0Þf ðx;y;zÞ ¼

x0^z0y0

x^zy¼ ðx0x _

x0yz Þ ^ ð z0

xy0 _z0yzy0ÞP d ^ d2 ¼ d2. Thus ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼

1 ^ f ðx;y;zÞf ðx0 ;y0 ;z0Þ ^

f ðx0 ;y0;z0Þf ðx;y;zÞ P d2:

If d = 0, then there exist x ¼ y ¼ z ¼ 0; y0 ¼ x0 ¼ z0 ¼ 1 such that Df ðdÞ ¼ d2 ¼ 0.

From the above discussion, we conclude that Mf ðdÞ ¼ d2.

(2)(c) Suppose d 2 ð0; 1�; x � y ¼ xy, x!GOy ¼ 1; if x ¼ 0;yx ^ 1; if x 6¼ 0;

�x � ðdÞx0, y � ðdÞy0, and z � ðdÞz0. Then

by Lemma 2.2, we conclude that dt0 6 t 6 1 ^ t0d ; dt 6 t0 6 1 ^ t

d ; 8t 2 fx; y; zg.

If one of x; y and z is equal to zero, then the value of ½f ðx; y; zÞ � f ðx0; y0; z0Þ� is either equal to 1 or not lessthan d. Thus the following discussion is studied under the condition that all of x; y; z; x0; y0 and z0 are belong toð0; 1�. There are two cases to be considered here.


Case 1: y 6 z. So f ðx; y; zÞ ¼ x. There are two sub-cases to be discussed in this case: y0 6 z0, or z0 6 y0. Ify0 6 z0, then f ðx0; y0; z0Þ ¼ x0. Thus ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ ½x � x0�P d: If z0 6 y0,then f ðx0; y0; z0Þ ¼ x0z0

y0 . It

follows that f ðx;y;zÞf ðx0;y0;z0Þ ¼ x

x0z0y0¼ xy0

x0z0 P d, and f ðx0;y0;z0Þf ðx;y;zÞ ¼

x0z0y0

x ¼ x0z0xy0 P

x0z0xðyd^1Þ ¼

x0z0dxy _ x0z0

x P x0z0dxz _ x0z0

x P d3 _ dz0 P d3.

Thus ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ 1 ^ f ðx;y;zÞf ðx0;y0;z0Þ ^

f ðx0;y0;z0Þf ðx;y;zÞ P d3. If we choose x ¼ y0 ¼ 1; x0 ¼ y ¼ z ¼ d; z0 ¼ d2,

then ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ d3.Case 2: z 6 y. So f ðx; y; zÞ ¼ xz

y . There are two sub-cases to be discussed: y0 6 z0, or z0 6 y0. If y0 6 z0, thenf ðx0; y0; z0Þ ¼ x0. It follows that f ðx;y;zÞ

f ðx0;y0;z0Þ ¼xzy

x0 ¼ xzx0y P dxz0

x0y P dxy0

x0y P d3, f ðx0;y0;z0Þf ðx;y;zÞ ¼ x0

xzy¼ x0y

xz P x0x P d. Therefore,

½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ 1 ^ f ðx;y;zÞf ðx0;y0;z0Þ ^

f ðx0;y0;z0Þf ðx;y;zÞ P d3:

If z0 6 y0, then f ðx0; y0; z0Þ ¼ x0z0y0 . It follows that f ðx;y;zÞ

f ðx0;y0;z0Þ ¼xzy

x0z0y0¼ xzy0

yx0z0 P d3, and f ðx0;y0;z0Þf ðx;y;zÞ ¼

x0z0y0xzy¼ x0z0y

xzy0 P d3.

Therefore, ½f ðx; y; zÞ � f ðx0; y0; z0Þ� ¼ 1 ^ f ðx;y;zÞf ðx0;y0;z0Þ ^

f ðx0;y0;z0Þf ðx;y;zÞ P d3.

If d = 0, then there exist x ¼ y ¼ z ¼ 0; y0 ¼ x0 ¼ z0 ¼ 1 such that Df ðdÞ ¼ d3 ¼ 0.From the above discussion, we conclude that Mf ðdÞ ¼ d3. h

Now we go to estimate the perturbation parameters of multidimensional fuzzy reasoning.

Lemma 2.3. Let hðx1; . . . ; xn; yÞ ¼ x1 � � � � � xn ! y ¼ x! y. Where x ¼ x1 � � � � � xn. Let 4�nðdÞ ¼ k. Then theresults of 4hðdÞ are contained in the following table.

! Lukasiewicz implication Mamdani implication Kleene-Dienes implication Goguen implication

4hðdÞ ðkþ d� 1Þ _ 0 k ^ d k ^ d dk

Proof

(1) If x! y ¼ 1 ^ ð1� xþ yÞ, x � ðkÞx0 and y � ðdÞy0, then by Lemma 1.2, we have½x! y � x0 ! y0� ¼ 1� jx! y � x0 ! y0j ¼ 1� j1 ^ ð1� xþ yÞ � 1 ^ ð1� x0 þ y 0ÞjP ð1� jx0 � xþ y�y0jÞ _ 0 P ð1� jx� x0j � jy � y0jÞ _ 0 P ðkþ d� 1Þ _ 0. Let x ¼ y ¼ k, x0 ¼ 1, y0 ¼ ðk� 1þ dÞ _ 0. Thenthe equality holds. Hence MhðdÞ ¼ ðkþ d� 1Þ _ 0.

(2) If x! y ¼ x ^ y, x � ðkÞx0 and y � ðdÞy 0, then by Lemma 2.2(1), it follows that x P k, x0 P k, y P d,y0 P d, and ½x! y � x0 ! y0� ¼ x ^ y ^ x0 ^ y0 P k ^ d. Let x ¼ x0 ¼ k, y ¼ y0 ¼ d. Then the equalityholds. Hence MhðdÞ ¼ k ^ d.

(3) If x! y ¼ ð1� xÞ _ y, x � ðkÞx0 and y � ðdÞy0, then by Lemma 2.2(1), it follows that ð1� xÞ _ x0 P k,ð1� x0Þ _ x P k, ð1� yÞ _ y0 P d and ð1� y 0Þ _ y P d, and thus ½x! y � x0 ! y0� ¼ ½½x ^ ð1� yÞ�_ð1� x0Þ _ y0� ^ ½½x0 ^ ð1� y 0Þ� _ ð1� xÞ _ y� ¼ ½x _ ð1� x0Þ _ y 0� ^ ½ð1� yÞ _ ð1� x0Þ _ y0� ^ ½x0 _ ð1� xÞ_y� ^ ½ð1� y0Þ _ ð1� xÞ _ y�P ½x _ ð1� x0Þ� ^ ½x0 _ ð1� xÞ� ^ ½y _ ð1� y 0Þ� ^ ½y 0 _ ð1� yÞ�P k ^ d. Fork 6 d, let x ¼ y 0 ¼ 0; x0 ¼ 1� k, and y ¼ 1� d. Then the equality holds. For k > d, let x ¼ x0 ¼ y0 ¼ 1and y ¼ d. Then the equality holds. Hence MhðdÞ ¼ k ^ d.

(4) If x! GOy ¼ 1; if x ¼ 0;yx ^ 1; if x 6¼ 0;

�x � ðkÞx0 and y � ðdÞy0 for some k 6¼ 0 and d 6¼ 0, then by Lemma 2.2(3),

we have kx0 6 x 6 x0

k ^ 1; dy0 6 y 6 y0

d ^ 1; kx 6 x0 6 xk ^ 1; dy 6 y0 6 y

d ^ 1, hðx; yÞ ¼ x! y ¼1; if x ¼ 0;yx ^ 1; if x 6¼ 0:

�hðx0; y0Þ ¼ 1; if x0 ¼ 0;

y0

x0 ^ 1; if x0 6¼ 0:

�If any element of x; x0; y and y 0 is equal to 0, then

½hðx; yÞ � hðx0; y 0Þ� ¼ 1; if x; x0; y; y 0 2 ð0; 1�, then there are four following cases:

Case 1: x 6 y; x0 6 y0: Then hðx; yÞ ¼ hðx0; y0Þ ¼ 1; ½hðx; yÞ � hðx0; y 0Þ� ¼ 1.Case 2: x 6 y; x0 P y0. Then hðx; yÞ ¼ 1; hðx0; y 0Þ ¼ y0

x0 ; ½hðx; yÞ � hðx0; y0Þ� ¼ ð1! y0

x0Þ ^ ðy0

x0 ! 1Þ ¼ y0

x0 Pdyx0 P

dxx0 P dk. If we choose x0 ¼ 1; x ¼ y ¼ k; y0 ¼ dk, then ½hðx; yÞ � hðx0; y0Þ� ¼ dk.

Case 3: x P y; x0 6 y 0. Then hðx; yÞ ¼ yx ; hðx0; y 0Þ ¼ 1; ½hðx; yÞ � hðx0; y0Þ� ¼ ðyx! 1Þ ^ ð1! y

xÞ ¼yx P

dy0

x P dx0

x P dk.Case 4: x P y; x0 P y 0. Then hðx; yÞ ¼ y

x ; hðx0; y0Þ ¼y0

x0. It follows that ½hðx; yÞ � hðx0; y0Þ�ðyx !¼y0

x0Þ ^ ðy0

x0 !yxÞ. Suppose y

x 6y0

x0, then ½hðx; yÞ � hðx0; y0Þ� ¼yxy0x0¼ yx0

xy0 P dk.


If d = 0 or k = 0, then there also exist y0; x0; x and y such that DhðdÞ ¼ dk ¼ 0.Therefore DhðdÞ ¼ dk. h

The computation of 4�nðdÞ are presented as follows:

Lemma 2.4. Let gðx1; . . . ; xnÞ ¼ x1 � � � � � xn.

(1) Suppose � ¼ ^. If ! is Lukasiewicz, Goguen, Mamdani or Kleene-Dienes implication, then 4gðdÞ ¼ d.

(2) Suppose * is the product operator. If ! is Lukasiewicz, Goguen, Mamdani or Kleene-Dienes implication,

then 4gðdÞ ¼ dn.

(3) Suppose * is Lukasiewicz t-norm operator. If ! is Lukasiewicz, Mamdani or Kleene-Dienes implication,then 4gðdÞ ¼ 0 _ ðnd� nþ 1Þ.

Proof

(1) Suppose x � y ¼ x ^ y and x! y ¼ 1 ^ ð1� xþ yÞ. Assume that ½xi � x0i�P dði ¼ 1; . . . ; nÞ. Then byLemmas 1.1 and 1.2, it follows that jxi � x0ij 6 1� dði ¼ 1; . . . ; nÞ, and ½x1 � x2 � � � � � xn � x01�x02 � � � � � x0n� ¼ 1� jx1 � x2 � � � � � xn � x01 � x02 � � � � � x0nj ¼ 1� jx1 ^ x2 ^ � � � ^ xn � x01 ^ x02 ^ � � � ^ x0njP 1�Wn

i¼1jxi � x0ijP 1� ð1� dÞ ¼ d. Where, let xi ¼ 1; x0i ¼ d; ði ¼ 1; . . . ; nÞ. Then ½x1 � x2 � � � � � xn �x01 � x02 � � � � � x0n� ¼ d. Hence 4gðdÞ ¼ d.

Suppose a! b ¼ 1; if a ¼ 0;ba ^ 1; if a 6¼ 0:

�½xi � yi�P d ði ¼ 1; 2; . . . ; nÞ; d 2 ð0; 1�. By Lemma 2.2(3), it is

equivalent to say that dxi 6 yi 6xid ^ 1 ði ¼ 1; 2; . . . ; nÞ.

If any element of xi and yi is equal to 0, then ½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ 1.If xi; yi 2 ð0; 1� ði ¼ 1; 2; . . . ; nÞ, then ½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ ð

Vni¼1

xi !Vni¼1

yiÞ^ðVni¼1

yi !Vni¼1

xiÞ. By the symmetry ofVni¼1

xi andVni¼1

yi, we could assume thatVni¼1

xi 6Vni¼1

yi.

ThenVni¼1

xi !Vni¼1

yi ¼ 1,Vni¼1

yi !Vni¼1

xi ¼

Vni¼1

xi

Vni¼1

yi

PVni¼1

xiyi

P d. So ½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ�P d. If

we choose xi ¼ d; yi ¼ 1 ði ¼ 1; 2; . . . ; nÞ, then ½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ d.If d = 0, then there exist xi and yi ði ¼ 1; . . . ; nÞ, such that ½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ d ¼ 0.From the above discussion, we know that MgðdÞ ¼ d.If x! y ¼ x ^ y, then by Lemma 2.2(2),½xi � yi�P d if and only if xi P d and yi P dði ¼ 1; 2; . . . ; nÞ. Then ½

Vni¼1xi �

Vni¼1yi� ¼ ð

Vni¼1xiÞ^

ðVn

i¼1yiÞP d ^ d ¼ d. Let xi ¼ yi ¼ d, then the equality holds. Hence 4gðdÞ ¼ d.If ! is Kleene-Dienes implication, we prove the result by induction on n.Base: Let ½x1 � y1�P d and ½x2 � y2�P d, then we have ½x1 ^ x2 � y1 ^ y2� ¼½ðx1 ^ x2Þ ! ðy1 ^ y2Þ� ^ ½ðy1 ^ y2Þ ! ðx1 ^ x2Þ� ¼ ½ð1 � x1Þ _ ð1 � x2Þ _ ðy1 ^ ^y2Þ�½ð1 � y1Þ _ ð1 � y2Þ_ðx1^x2Þ�¼ ½ð1�x1Þ_ð1�x2Þ_y1�^ ½ð1�x1Þ_ð1�x2Þ_y2�^ ½ð1�y1Þ_ð1�y2Þ__x1�^ ½ð1�y1Þ_ð1�y2Þ_x2�P ½ð1� x1Þy1� ^ ½ð1� x2Þ _ y2� ^ ^½ð1� y1Þ _ x1� ^ ½ð1� y2Þ_ x2� ¼ ½x1 � y1� ^ ½x2 � y2�P d ^ d ¼ d.If d 6 1

2, we take x1 ¼ x2 ¼ 1� d, y1 ¼ y2 ¼ d, then the equality holds. If d P 1

2, we choose

x1 ¼ x2 ¼ y1 ¼ y2 ¼ d, then the equality holds.

Induction: Suppose it is valid for n ¼ k, i.e., if xi � ðdÞyiði 2 NÞ, then ½Vk

i¼1xi �Vk

i¼1yi�P d. It follows

that ½Vkþ1

i¼1 xi �Vkþ1

i¼1 yi� ¼ ½ð1�Vkþ1

i¼1 xiÞ _Vkþ1

i¼1 yi� ^ ½ðð1�Vkþ1

i¼1 yiÞ _Vkþ1

i¼1 xi� ¼ ½ð1�Vk

i¼1xiÞ _ ð1� xkþ1Þ_Vkþ1i¼1 yi�^ ½ð1�

Vki¼1yiÞ_ð1� ykþ1Þ_

Vkþ1i¼1 xi� ¼ ½ð1�

Vki¼1xiÞ_ð1�xkþ1Þ_

Vki¼1yi�^ ½ð1�

Vki¼1xiÞ_ð1�xkþ1Þ_

ykþ1� ^ ½ð1 �Vk

i¼1yiÞ _ ð1 � ykþ1Þ _Vk

i¼1xi� ^ ½ð1 �Vk

i¼1yiÞ _ ð1 � ykþ1Þ _ xkþ1�P ½ð1 �Vk

i¼1xiÞ _Vk

i¼1yi�^½ð1� xkþ1Þ_ ykþ1�^ ½ð1�

Vki¼1yiÞ_

Vki¼1xi� ^ ½ð1� ykþ1Þ_ xkþ1� ¼ ½

Vki¼1xi�

Vki¼1yi�^ ½xkþ1^ ykþ1�P d^d¼ d.

For d 6 12,let

Vki¼1xi ¼ xkþ1 ¼ 1� d and

Vki¼1yi ¼ ykþ1 ¼ d. Then all of the above equalities hold. For

d > 12, let

Vki¼1xi ¼ xkþ1 ¼

Vki¼1yi ¼ ykþ1 ¼ d. Then all of the above equalities hold.

By the induction we conclude that 4gðdÞ ¼ d.


(2) Suppose x! y ¼ 1 ^ ð1� xþ yÞ, x � y ¼ xy and ½xi � yi�P dði ¼ 1; . . . ; nÞ. Then the last inequalityimplies jxi � yij 6 1� dði ¼ 1; . . . ; nÞ by Lemma 1.2. It is no generality to assume that x1 � x2 Py1 � y2. We prove the result by induction on n.Base: ½x1 � x2 � y1 � y2� ¼ 1� jx1x2 � y1y2j ¼ 1� x1x2 þ y1y2 P 1� x1x2 þ ðx1 � 1þ dÞðx2 � 1þ dÞ ¼ 1�x1x2 þ x1x2 � ð1� dÞðx1 þ x2Þ þ ð1� dÞ2 P 1� 2ð1� dÞ þ ð1� dÞ2 ¼ d2 Let x1 ¼ x2 ¼ 1; y1 ¼ y2 ¼ d.Then all of the above equalities hold.Induction: Suppose ½x1 � x2 � � � � � xk � y1 � y2 � � � � � yk�P dk and x1 � x2 � � � � � xn P y1 � y2 � � � � �ynð8n 2 NÞ. Then ½x1 � x2 � � � � � xkþ1 � y1 � y2 � � � � � ykþ1� ¼ 1� x1x2 � � � xkþ1 þ y1y2 � � � ykþ1 P 1� x1x2 � � �xkþ1þ ðx1x2 � � �xk � 1þ dkÞðxkþ1� 1þ dÞ ¼ 1� x1x2 � � �xkþ1þ x1x2 � � �xkþ1� ð1� dÞx1x2 � � �xk � ð1� dkÞxkþ1þð1� dkÞð1� dÞP 1� ð1� dkÞ � ð1� dÞ þ ð1� dkÞð1� dÞ ¼ dkþ1. Let xi ¼ 1; yi ¼ dði ¼ 1; . . . ; k þ 1Þ.Then all of the above equalities hold.By the induction, it follows that 4gðdÞ ¼ dn.

Suppose x � y ¼ xy; a! b ¼ 1; if a ¼ 0;ba ^ 1; if a 6¼ 0

�and ½xi � yi�P d; ði ¼ 1; 2; . . . ; nÞ; d 2 ð0; 1�. By Lemma

2.2(3), it is equivalent to say that dyi 6 xi 6yid ^ 1 ði ¼ 1; 2; . . . ; nÞ. If any element of xi and

yi ði ¼ 1; 2; . . . ; nÞ is equal to 0, then ½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ ½0 � 0� ¼ 1. Ifxi; yi 2 ð0; 1� ði ¼ 1; 2; . . . ; nÞ, then by the symmetry of xi and yi, we can assume that x1x2 � � �xn 6 y1y2 � � � yn.Thus, ½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ x1x2��xn

y1y2��ynP dn. Let xi ¼ d; yi ¼ 1 ði ¼ 1; 2; . . . ; nÞ. Then

½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ dn. If d = 0, then there exist xi and yi such that½gðx1; x2; . . . ; xnÞ � gðy1; y2; . . . ; ynÞ� ¼ 0. So MgðdÞ ¼ dn.If a! b ¼ a ^ b, then by Lemma 2.2(1), ½xi � yi�P d if and only if xi P d and yi P d ði ¼ 1; 2; . . . ; nÞ. Itfollows that ½

Qni¼1xi �

Qni¼1yi� ¼ ð

Qni¼1xi ^ ð

Qni¼1yiÞP dn ^ dn ¼ dn. If we choose xi ¼ yi ¼

d ði ¼ 1; . . . ; nÞ, then the equality holds. Hence 4gðdÞ ¼ dn.If x! y ¼ ð1� xÞ _ y, x � ðdÞx0 and y � ðdÞy 0, then ½xy � x0y0� ¼ ðxy ! x0y0Þ ^ ðx0y0 ! xyÞ ¼½ð1� xyÞ _ x0y0� ^ ½ð1� x0y0Þ _ xy�. In this case, we prove the result by induction on n.Base: Since ½x � x0�P d; ½y � y0�P d, by Lemma 2.2(2), it is equivalent to say that ðx; x0Þ 2 A [ B andðy; y0Þ 2 C [ D, where A ¼ fðx; x0Þj1� x P x0; 1� x P d; 1� x0 P dg, B ¼ fðx; x0Þj1� x 6 x0; x P d;x0 P dg, C ¼ fðy; y 0Þj1� y P y0; 1� y P d; 1� y0 P dg, D ¼ fðy; y0Þj1� y 6 y0; y P d; y 0 P dg.If 1� xy P x0y 0, then we have to consider four cases:Case 1: ðx; x0Þ 2 A and ðy; y0Þ 2 D. Then ½xy � x0y 0� ¼ ð1� xyÞ ^ ð1� x0y 0ÞP ð1� xÞ ^ ð1� x0ÞP d. If wechoose x ¼ 1� d; x0 ¼ 0; y ¼ y0 ¼ 1, then the equality holds.Case 2: ðx; x0Þ 2 A and ðy; y0Þ 2 C. Then 1� xy P 1� ð1� dÞ2 ¼ 2d� d2, 1� x0y0 P1� ð1� dÞ2 ¼ 2d� d2. Hence ½xy � x0y0� ¼ ð1� xyÞ ^ ð1� x0y0ÞP 2d� d2. If we choose x ¼ y ¼ 1� d;x0 ¼ y0 ¼ 0, then the equality holds.Case 3: ðx; x0Þ 2 B and ðy; y 0Þ 2 C. Then 1� xy P 1� ð1� dÞ ¼ d, 1� x0y 0 P 1� ð1� dÞ ¼ d. Hence½xy � x0y0� ¼ ð1� xyÞ ^ ð1� x0y0ÞP d. If we choose x ¼ x0 ¼ 1; y ¼ 1� d; y0 ¼ 0, then the equalityholds.Case 4: ðx; x0Þ 2 B and ðy; y0Þ 2 D. Then d2

6 x0y0 6 1� xy 6 1� d2. So if d 2 ½0;ffiffi2p

2�, then

½xy � x0y0�P d2; if d 2 ðffiffi2p

2; 1�, then 1� xy < x0y0, which contradicts the assumption of 1� xy P x0y 0.

On the other hand, if 1� xy 6 x0y 0, then 1� x 6 x0 and 1� y 6 y0. By Lemma 2.2(2), we havex P d; x0 P d; y P d; y 0 P d. Hence ½xy � x0y 0� ¼ xy ^ x0y0 P d2 ^ d2 ¼ d2. Let x ¼ d; x0 ¼y0 ¼ 1; y ¼ d. Then the equality holds.In summary, we always have ½xy � x0y0�min ¼ d2.Induction: Suppose it is valid for n ¼ k. That is, if xi � ðdÞx0i ði ¼ 1; . . . ; kÞ, thenx1x2 � � � xk � ðdkÞx01x02 � � � x0k. By Lemma 2.2(2), we know that ½x1x2 � � � xk � x01x02 � � � x0k�P dk is equivalentto the conditions 1� x1x2 � � � xk P x01x02 � � � x0k, 1� x1x2 � � � xk P dk, and 1� x01x02 � � � x0k P dk or the condi-tions 1� x01x02 � � � x0k 6 x1x2 � � � xk; x1x2 � � � xk P dk and x01x02 � � � x0k P dk, the condition xkþ1 � ðdÞx0kþ1 isequivalent to the conditions 1� xkþ1 P x0kþ1; 1� xkþ1 P d; 1� x0kþ1 P d or the conditions 1� xkþ1 6

x0kþ1; xkþ1 P d; x0kþ1 P d. Note ½x1x2 � � � xkþ1 � x01x02 � � � x0kþ1� ¼ ½ð1� x1x2 � � � xkþ1Þ _ x01x02 � � � x0kþ1�^½ð1� x01x02 � � � x0kþ1Þ _ x1x2 � � � xkþ1�, if 1� x1x2 � � � xkþ1 P x01x02 � � � x0kþ1, we have to consider four cases here.Case 1: 1� x1x2 � � � xk P x01x02 � � � x0k; 1� x1x2 � � � xk P dk; 1� x01x02 � � � x0k P dk; 1� xkþ1 P x0kþ1; 1� xkþ1 P


d; 1� x0kþ1 P d. Then 1� x1x2 � � � xkþ1 P 1� ð1� dkÞð1� dÞ ¼ dk þ d� dkþ1, 1� x01x02 � � � x0kþ1 P1� ð1� dkÞð1� dÞ ¼ dk þ d� dkþ1. And thus ½x1x2 � � � xkþ1 � x01x02 � � � x0kþ1� ¼ ð1� x1x2 � � � xkþ1Þ^ð1� x01x02 � � � x0kþ1ÞP dk þ d� dkþ1. If we choose x1x2 � � � xk ¼ 1� dk; xkþ1 ¼ 1� d; x01x02 � � � x0k ¼ x0kþ1 ¼ 0,then the equality holds.Case 2: 1� x1x2 � � � xk P x01x02 � � � x0k; 1� x1x2 � � � xk P dk; 1� x01x02 � � � x0k P dk; 1� xkþ1 6 x0kþ1; xkþ1 Pd; x0kþ1 P d. Then ½x1x2 � � � xkþ1 � x01x02 � � � x0kþ1� ¼ ð1� x1x2 � � � xkþ1Þ ^ ð1� x01x02 � � � x0kþ1ÞP ð1� x1x2 � � �xkÞ ^ ð1� x01x02 � � � x0kÞP dk. If we choose x1x2 � � � xk ¼ 1� dk; xkþ1 ¼ x0kþ1 ¼ 1; x01x02 � � � x0k ¼ 0, then theequality holds.Case 3: 1� x1x2 � � � xk 6 x01x02 � � � x0k; x1x2 � � � xk P dk; x01x02 � � � x0k P dk; 1� xkþ1 P x0kþ1; 1� xkþ1 P d; 1�x0kþ1 P d. Then 1� x1x2 � � � xkþ1 P 1� ð1� dÞ ¼ d, 1� x01x02 � � � x0kþ1 P 1� ð1� dÞ ¼ d. And thus½x1x2 � � � xkþ1 � x01x02 � � � x0kþ1� ¼ ð1� x1x2 � � � xkþ1Þ ^ ð1� x01x02 � � � x0kþ1ÞP d. If we choose x1x2 � � � xk ¼x01x02 � � � x0k ¼ 1; xkþ1 ¼ 1� d; x0kþ1 ¼ 0, then the equality holds.Case 4: 1� x1x2 � � � xk 6 x01x02 � � � x0k; x1x2 � � � xk P dk; x01x02 � � � x0k P dk; 1� xkþ1 6 x0kþ1; xkþ1 P d; x0kþ1 P d.Then dkþ1

6 x01x02 � � � x0kþ1 6 1� x1x2 � � � xkþ1 6 1� dkþ1. So if d 2 ½0; 2� 1kþ1�, then ½x1x2 � � � xkþ1 �

x01x02 � � � x0kþ1� ¼ ð1� x1x2 � � � xkþ1Þ ^ ð1� x01x02 � � � x0kþ1ÞP dkþ1; else if d 2 ð2� 1kþ1; 1�, then

1� x1x2 � � � xkþ1 < x01x02 � � � x0kþ1, which contradicts the assumption of 1� x1x2 � � � xkþ1 P x01x02 � � � x0kþ1.On the other hand, if 1� x1x2 � � � xkþ1 6 x01x02 � � � x0kþ1, it follows that both inequalities 1� x1x2 � � � xk 6

x01x02 � � � x0k and 1� xkþ1 6 x0kþ1 hold, which leads to x1x2 � � � xk P dk; x01x02 � � � x0k P dk; xkþ1 P d andx0kþ1 P d by Lemma 2.2(2). So ½x1x2 � � � xkþ1 � x01x02 � � � x0kþ1� ¼ x1x2 � � � xkþ1 ^ x01x02 � � � x0kþ1 P x1x2 � � � xk^x01x02 � � � x0k P dkd ¼ dkþ1. Let x1x2 � � � xk ¼ dk, x01x02 � � � x0k ¼ 1, xkþ1 ¼ d and x0kþ1 ¼ 1. Then all of the aboveequalities hold.In summary, we conclude that ½x1x2 � � � xkþ1 � x01x02 � � � x0kþ1�min ¼ dkþ1.By the induction, it follows that 4gðdÞ ¼ dn.

(3) Suppose x � y ¼ 0 _ ðxþ y � 1Þ, x! y ¼ 1 ^ ð1� xþ yÞ and ½xi � yi�P dði ¼ 1; . . . ; nÞ. Then the lastinequality implies jxi � yij 6 1� dði ¼ 1; . . . ; nÞ by Lemma 1.2. We prove the result by induction on n.Base: ½x1 � x2 � y1 � y2� ¼ 1� jx1 � x2 � y1 � y2j ¼ 1� j0 _ ðx1 þ x2 � 1Þ � 0 _ ðy1 þ y2 � 1ÞjP ð1� jx1�y1 þ x2 � y2jÞ _ 0 P ð1� jx1 � y1j � jx2 � y2jÞ _ 0 P ð1� 2ð1� dÞÞ _ 0 ¼ ð2d� 1Þ _ 0. Let x1 ¼ x2 ¼ 1,y1 ¼ y2 ¼ d. Then all of the above equalities hold.Induction: Suppose ½x1 � x2 � � � � xk � y1 � y2 � � � � yk�P ðkd� k þ 1Þ _ 0. Then ½x1 � x2 � � � � xkþ1 �y1 � y2 � � � � ykþ1� ¼ 1�jx1 � x2 � � � � xkþ1� y1 � y2 � � � � ykþ1j ¼ 1�j0_ðx1 � x2 � � � � xkþ xkþ1�1Þ�0_ðy1 � y2 � � ��yk þ ykþ1 � 1ÞjP ð1� jx1 � x2 � � � � xk þ xkþ1 � y1 � y2 � � � � yk � ykþ1jÞ _ 0 P ð1� jx1 � x2 � � � � xk � y1 � y2 � � � �ykj � jxkþ1 � ykþ1jÞ _ 0 P ððkd� k þ 1Þ _ 0� ð1� dÞÞ _ 0 ¼ ððk þ 1Þd� ðk þ 1Þ þ 1Þ _ 0. Let x1 � x2 � � � �xk ¼ xkþ1 ¼ 1, y1 � y2 � � � � yk ¼ ðkd� k þ 1Þ _ 0, ykþ1 ¼ d. Then all of the above equalities hold.By the induction it follows that 4gðdÞ ¼ 0 _ ðnd� nþ 1Þ.Suppose x! y ¼ x ^ y. Let xi � ðdÞx0iði ¼ 1; . . . ; nÞ. It follows that xi P d and x0i P d by Lemma 2.2(1).Then ½x1 � x2 � x01 � x02� ¼ ððx1 þ x2 � 1Þ _ 0Þ ^ ððx01 þ x02� 1Þ _ 0ÞP ð2d� 1Þ _ 0.Suppose x1 � x2 � � � � � xk � ðððkd� k þ 1Þ _ 0ÞÞx01 � x02 � � � � � x0k. Then ½x1 � x2 � � � � � xkþ1 � x01 � x02 � � � ��x0kþ1� ¼ ððx1 � x2 � � � � � xk þ xkþ1 � 1Þ _ 0Þ ^ ððx01 � x02 � � � � � x0k þ x0kþ1 � 1Þ _ 0ÞP ððkd� k þ 1Þ _ 0þd� 1Þ _ 0 ¼ ððk þ 1Þd� ðk þ 1Þ þ 1Þ _ 0. Let xi ¼ x0i ¼ dði ¼ 1; . . . ; nÞ. Then the equality holds.By the induction it follows that 4gðdÞ ¼ ðnd� nþ 1Þ _ 0.Suppose x! y ¼ ð1� xÞ _ y. We prove the result by induction on n.Base: Since ½x � x0�P d; ½y � y0�P d, by Lemma 2.2(2), it is equivalent to the conditions ðx; x0Þ 2 A [ Band ðy; y 0Þ 2 C [ D, where A ¼ fðx; x0Þj1� x P x0; 1� x P d; 1� x0 P dg, B ¼ fðx; x0Þj1� x 6x0; x P d; x0 P dg,C ¼ fðy; y0Þj1� y P y0; 1� y P d; 1� y0 P dg, D ¼ fðy; y 0Þj1� y 6 y0; y P d; y0 P dg.Note ½x � y � x0 � y0� ¼ ðx � y ! x0 � y0Þ ^ ðx0 � y0 ! x � yÞ ¼ ½ð1� x � yÞ _ x0� y0� ^ ½ð1� x0 � y 0Þ _ x � y�. If1� x � y P x0 � y0, we have to consider four cases:Case 1: ðx; x0Þ 2 A and ðy; y 0Þ 2 D. Then ½x � y � x0 � y0� ¼ ð1� x � yÞ ^ ð1� x0 � y0ÞP 1 ^ ð2� x� yÞ^ð2� x0 � y 0ÞP ð1� xÞ ^ ð1� x0ÞP d. Let x ¼ 1� d; x0 ¼ 0; y ¼ y 0 ¼ 1. Then all of the above equalitieshold.Case 2: ðx; x0Þ 2 A and ðy; y0Þ 2 C. Then 1� x � y ¼ 1 ^ ð2� x� yÞP 1 ^ 2d, and 1� x0 � y0 ¼ 1^ð2� x� yÞP 1 ^ 2d. Hence ½x � y � x0 � y 0� ¼ ð1� x � yÞ ^ ð1� x0 � y0ÞP 2d ^ 1. Let x ¼ y ¼ 1�d; x0 ¼ y0 ¼ 0. Then all of the above equalities hold.


Case 3: ðx; x0Þ 2 B and ðy; y0Þ 2 C. Then 1� x � y ¼ 1 ^ ð2� x� yÞP 1 ^ ð1� 1þ dÞ ¼ d, and1� x0 � y 0 ¼ 1 ^ ð2� x0 � y0ÞP 1 ^ ð1� 1þ dÞ ¼ d. So ½x � y � x0 � y0�P d. Let x ¼ x0 ¼ 1; y ¼ 1� d;y0 ¼ 0. Then all of the above equalities hold.Case 4: ðx; x0Þ 2 B and ðy; y0Þ 2 D. Then 1 ^ ð2� 2dÞP 1 ^ ð2� x� yÞP 1� x � y P x0 � y0 P0 _ ðx0 þ y0 � 1ÞP 0 _ ð2d� 1Þ, and 1 ^ ð2� 2dÞP 1 ^ ð2� x0 � y 0ÞP 1� x0 � y0 P x � y P 0_ðxþ y � 1ÞP 0 _ ð2d� 1Þ. Hence if d 2 ½0; 3

4�, then ½x � y � x0 � y0� ¼ ð1� x � yÞ ^ ð1� x0 � y 0ÞP

0 _ ð2d� 1Þ. Let x ¼ x0 ¼ 1; y ¼ 1� d; y0 ¼ 0. Then the equality holds; for d 2 ð34; 1�, the condition of

1� x0 � y 0 P x � y is not true.On the other hand, if 1� x � y 6 x0 � y0, then 1� x 6 x0 and 1� y 6 y 0. By Lemma 2.2(2), we havex P d; x0 P d; y P d; y0 P d. Hence ½x � y � x0 � y 0� ¼ ðx � yÞ ^ ðx0 � y0ÞP 0 _ ½ðxþ y � 1Þ^ðx0 þ y0 � 1Þ�P 0 _ ð2d� 1Þ. Let x ¼ d; x0 ¼ y 0 ¼ 1; y ¼ d. Then all of the above equalities hold.In summary, it follows that ½x � y � x0 � y 0�min ¼ 0 _ ð2d� 1Þ.Induction: Suppose that the result is valid for n ¼ k. That is, if xi � ðdÞx0i ði ¼ 1; . . . ; kÞ, then x1 � x2�� xk � ð0 _ ðkd� k þ 1ÞÞx01 � x02 � � � � � x0k. By Lemma 2.2(2), we see that ½x1 � x2 � � � � � xk � x01 � x02�� x0k�P 0 _ ðkd� k þ 1Þ is equivalent to ‘‘1� x1 � x2 � � � � � xk P x01 � x02 � � � � � x0k, 1� x1� x2�� xk P 0 _ ðkd� k þ 1Þ, and 1� x01 � x02 � � � � � x0k P 0 _ ðkd� k þ 1Þ’’ or ‘‘1� x01 � x02 � � � � � x0k 6x1 � x2 � � � � � xk; x1 � x2 � � � � � xk P 0 _ ðkd� k þ 1Þ and x01 � x02 � � � � � x0k P 0 _ ðkd� k þ 1Þ’’. We have½x1 � x2 � � � � � xkþ1 � x01 � x02 � � � � � x0kþ1� ¼ ½ð1� x1 � x2 � � � � � xkþ1Þ _ x01 � x02 � � � � �x0kþ1� ^ ½ð1� x01 � x02�� x0kþ1Þ _ x1 � x2 � � � � � xkþ1�.If 1� x1 � x2 � � � � � xkþ1 P x01 � x02 � � � � � x0kþ1, then we have to consider four cases:Case 1: 1� x1 � x2 � � � � � xk P x01 � x02 � � � � � x0k; 1� x1 � x2 � � � � � xk P 0 _ ðkd� k þ 1Þ; 1� x01 � x02 � � � � �x0k P 0 _ ðkd� k þ 1Þ; 1� xkþ1 P x0kþ1; 1� xkþ1 P d; 1� x0kþ1 P d. Then 1� x1 � x2 � � � � � xkþ1 ¼ 1� 0_ðx1 � x2 � � � � � xk þ xkþ1 � 1Þ ¼ 1 ^ ð2� x1 � x2 � � � � � xk � xkþ1ÞP 1 ^ ð0 _ ðkd� k þ 1Þ þ dÞ ¼ d _ ð1^ððk þ 1Þd� k þ 1ÞÞ, 1� x01 � x02 � � � � � x0kþ1 ¼ 1� 0 _ ðx01 � x02 � � � � � x0k þ x0kþ1 � 1Þ ¼ 1 ^ ð2� x01 � x02 � � � ��x0k � x0kþ1ÞP 1 ^ ð0 _ ðkd� k þ 1Þ þ dÞ ¼ d _ ð1 ^ ððk þ 1Þd� k þ 1ÞÞ.And thus ½x1 � x2 � � � � x�kþ1 � x01 � x02 � � � � � x0kþ1� ¼ ð1� x1 � x2 � � � � � xkþ1Þ ^ ð1� x01 � x02 � � � � � x0kþ1ÞPd _ ð1 ^ ððk þ 1Þd� k þ 1ÞÞ. Let x1 � x2 � � � � � xk ¼ 1� 0 _ ðkd� k þ 1Þ; xkþ1 ¼ 1� d andx01 � x02 � � � � � x0k ¼ x0kþ1 ¼ 0. Then all of the above equalities hold.Case 2: 1� x1 � x2 � � � � � xk P x01 � x02 � � � � � x0k; 1� x1 � x2 � � � � � xk P 0 _ ðkd� k þ 1Þ; 1� x01 � x02 � � � ��x0k P 0 _ ðkd� k þ 1Þ; 1� xkþ1 6 x0kþ1; xkþ1 P d; x0kþ1 P d.Then ½x1 � x2 � � � � � xkþ1 � x01 � x02 � � � � � x0kþ1� ¼ ð1� x1 � x2 � � � � � xkþ1Þ ^ ð1� x01 � x02 � � � � � x0kþ1Þ ¼ 1^ð2� x1 � x2 � � � � � xk � xkþ1Þ ^ ð2� x01 � x02 � � � � � x0k � x0kþ1ÞP 1 ^ ð1� x1 � x2 � � � � � xkÞ ^ ð1� x01 � x02�� x0kÞP 1 ^ ð0_ ðkd� k þ 1ÞÞ. Let x1 � x2 � � � � � xk ¼ 1� 0 _ ðkd� k þ 1Þ; xkþ1 ¼ x0kþ1 ¼ 1 andx01 � x02 � � � � � x0k ¼ 0. Then all of the above equalities hold.Case 3: 1� x1 � x2 � � � � � xk 6 x01 � x02 � � � � � x0k; x1 � x2 � � � � � xk P 0 _ ðkd� k þ 1Þ; x01 � x02 � � � � � x0k P0 _ ðkd� k þ 1Þ; 1� xkþ1 P x0kþ1; 1� xkþ1 P d; 1� x0kþ1 P d.Then ½x1 � x2 � � � � � xkþ1 � x01 � x02 � � � � �x0kþ1� ¼ ð1 � x1 � x2 � � � � � xkþ1Þ ^ ð1 � x01 � x02 � � � � � x0kþ1Þ ¼ 1 ^ ð2 � x1 � x2 � � � � � xk � xkþ1Þ ^ ð2 � x01�x02 � � � � � x0k � x0kþ1ÞP 1 ^ ð1� xkþ1Þ ^ ð1� x0kþ1ÞP d. Let x1 � x2 � � � � � xk ¼ x01 � x02 � � � � � x0k ¼1; xkþ1 ¼ 1� d; and x0kþ1 ¼ 0: Then all of the above equalities hold.Case 4: 1� x1 � x2 � � � � � xk 6 x01 � x02 � � � � � x0k; x1 � x2 � � � � � xk P 0 _ ðkd� k þ 1Þ; x01 � x02 � � � � � x0k P0 _ ðkd� k þ 1Þ; 1� xkþ1 6 x0kþ1; xkþ1 P d; x0kþ1 P d. Then 1� x1 � x2 � � � � � xkþ1 P x01 � x02 � � � � �x0kþ1 ¼ 0 _ ðx01 � x02 � � � � � x0k þ x0kþ1 � 1ÞP 0 _ ð0 _ ðkd� k þ 1Þ þ d� 1Þ ¼ 0 _ ððk þ 1Þd� ðk þ 1Þ þ 1Þ,1� x01 � x02 � � � � � x0kþ1 P x1 � x2 � � � � � xkþ1 ¼ 0 _ ðx1 � x2 � � � � � xk þ xkþ1 � 1ÞP 0 _ ð0 _ ðkd� k þ 1Þþd� 1Þ ¼ 0 _ ððk þ 1Þd� ðk þ 1Þ þ 1Þ, and thus ½x1 � x2 � � � � � xkþ1 � x01 � x02 � � � � � x0kþ1� ¼ð1� x1 � x2 � � � � � xkþ1Þ ^ ð1� x01 � x02 � � � � � x0kþ1ÞP 0 _ ððk þ 1Þd� ðk þ 1Þ þ 1Þ.On the other hand, if 1� x1 � x2 � � � � � xkþ1 6 x01 � x02 � � � � � x0kþ1, it follows that both1� x1 � x2 � � � � � xk 6 x01 � x02 � � � � � x0k and 1� xkþ1 6 x0kþ1 hold, which leads tox1 � x2 � � � � � xk P 0 _ ðkd� k þ 1Þ; x01 � x02 � � � � � x0k P 0 _ ðkd� k þ 1Þ; xkþ1 P d and x0kþ1 P d byLemma 2.2(2).So x1 � x2 � � � � � xkþ1 ¼ 0 _ ðx1 � x2 � � � � � xk þ xkþ1 � 1ÞP 0 _ ð0 _ ðkd� k þ 1Þ þ d� 1Þ ¼ 0_ððk þ 1Þd� ðk þ 1Þ þ 1Þ, x01 � x02 � � � � � x0kþ1 ¼ 0 _ ðx01 � x02 � � � � � x0k þ x0kþ1 � 1ÞP 0 _ ð0 _ ðkd� k þ 1Þþd� 1Þ ¼ 0 _ ððk þ 1Þd� ðk þ 1Þ þ 1Þ, ½x1 � x2 � � � � � xkþ1 � x01 � x02 � � � � � x0kþ1� ¼ ðx1 � x2 � � � � � xkþ1Þ^


ðx01 � x02 � � � � � x0kþ1ÞP 0 _ ððk þ 1Þd� ðk þ 1Þ þ 1Þ. Let x1 � x2 � � � � � xk ¼ 0 _ ðkd� k þ 1Þ, x01 � x02 � � � ��x0k ¼ 1, xkþ1 ¼ d and x0kþ1 ¼ 1. Then all of the above equalities hold.In summary, it can be concluded that ½x1 � x2 � � � � � xkþ1 � x01 � x02 � � � � � x0kþ1�min ¼ 0 _ ððk þ 1Þd� ðk þ 1Þ þ 1Þ.By the induction it follows that 4gðdÞ ¼ 0 _ ðnd� nþ 1Þ. h

Theorem 2.2. Let f ðx01; . . . ; x0n; x1; . . . ; xn; yÞ ¼ ðx01�0 � � � x0nÞ � ½ðx1�0 � � � �0xnÞ ! y�, hðx1; . . . ; xn; yÞ ¼ ðx1�0 � � ��0xnÞ ! y, 4�0nðdÞ ¼ g and 4hðdÞ ¼ k. We have the following statements:

(1) Suppose * is the minimum conjunction and! is the Lukasiewicz, Goguen, Mamdani or Kleene-Dienes impli-

cation. Then 4f ðdÞ ¼ g ^ k;(2) Suppose * is the product conjunction and! is the Lukasiewicz, Goguen, Mamdani or Kleene-Dienes impli-

cation. Then 4f ðdÞ ¼ gk;

(3) Suppose * is the Lukasiewicz t-norm and ! is the Lukasiewicz, Mamdani or Kleene-Dienes implication.

Then 4f ðdÞ ¼ ðgþ k� 1Þ _ 0.

Proof. The course of proof is similar to that of Theorem 2.1. h

Now we have presented the best perturbation parameters of fuzzy reasoning according to Theorems 2.1 and2.2.

3. Robustness of fuzzy modus tollens

Fuzzy modus tollens (MT) comes in the following format:

Premise 1: If x is A, then y is B;Premise 2: y is B0;Conclusion: x is A0.

That is, A0 ¼ B0 � ðA! BÞ, where A0ðxÞ ¼W

y2Y ½B0ðyÞ � ðAðxÞ ! BðyÞÞ�, for any x 2 X .Obviously, the discussion of the robustness of the fuzzy MP rule in case n = 1 can apply to that of fuzzy MT

rule, as long as we note that A0 and B0 are to be positioned in place of B0 and A0 in the fuzzy MP rule. Let

4A0 ðdÞ ¼ inff½A01 � A02�jA1 � ðdÞA2;B1 � ðdÞB2;B01 � ðdÞB02g
denote the robustness measure of the fuzzy MT rule. Where, A0i ¼ B0i � ðAi ! BiÞ ði ¼ 1; 2Þ: Then Theorem 2.1is still valid for fuzzy MT. We present it in the following theorem.
Theorem 3.1. For the fuzzy MT rule, the following statements are valid:

(1) Suppose * is the minimum conjunction. Then, (a) if ! is the Lukasiewicz implication, then 4A0 ðdÞ ¼ð2d� 1Þ _ 0; (b) if! is the Mamdani or Kleene-Dienes implication, then4A0 ðdÞ ¼ d; (c) if! is the Goguen

implication, then 4A0 ðdÞ ¼ d2;

(2) Suppose * is the product conjunction. Then, (a) if ! is the Lukasiewicz implication, then 4A0 ðdÞ ¼ð2d2 � dÞ _ 0; (b) if! is the Mamdani or Kleene-Dienes implication, then 4A0 ðdÞ ¼ d2; (c) if! is the Gog-

uen implication, then 4A0 ðdÞ ¼ d3;

(3) Suppose * is the Lukasiewicz conjunction. Then, (a) if ! is the Lukasiewicz implication, then 4A0 ðdÞ ¼ð3d� 2Þ _ 0; (b) if ! is the Mamdani or Kleene-Dienes implication, then 4A0 ðdÞ ¼ ð2d� 1Þ _ 0.

4. Robustness of fuzzy reasoning machine

Adding the number of rules to the multidimensional fuzzy systems, we conclude multidimensional multi-rules fuzzy reasoning systems. The general form of the fuzzy reasoning is given as follows:


Premise 1: If x1 is A11 and x2 is A12 and � � � xn is A1n, then y is B1;. . .If x1 is Am1 and x2 is Am2 and � � � xn is Amn, then y is Bm;Premise 2: x1 is A01 and x2 is A02 and � � � xn is A0n;Conclusion: y is B0.In general, there are four different algorithms of fuzzy reasoning according to multidimensional multi-rules

fuzzy reasoning systems. It is stated as follows [6,7]:

Algorithm 1: B01 ¼Sm

i¼1A0 � ðAi ! BiÞ, Algorithm 2: B02 ¼ A0 �Sm

i¼1ðAi ! BiÞ.Algorithm 3: B03 ¼

Tmi¼1A0 � ðAi ! BiÞ; Algorithm 4: B04 ¼ A0 �

Tmi¼1ðAi ! BiÞ,

where [ and \ are s-norm and t-norm respectively, and ! is an implication operator.Now we propose the concept of robustness of fuzzy reasoning machine.

Definition 4.1. For any Aij;A0ij;A0j;A00j 2 F ðX jÞ, Bi;B0;B00 2 F ðY Þ ði ¼ 1; . . . ;m; j ¼ 1; . . . ; nÞ, and d 2 ½0; 1�, let

Ri ¼ Ci ! Bi, R0i ¼ C0i ! B0i, Ci ¼Qn

j¼1Aij, C0i ¼Qn

j¼1A0ij, C0 ¼Qn

j¼1A0j, C00 ¼Qn

j¼1A00j , B0ðyÞ ¼Sm

i¼1DiðyÞ,B00ðyÞ ¼

Smi¼1D0iðyÞ, DiðyÞ ¼

WxC0ðxÞ � Riðx; yÞ, D0iðyÞ ¼

WxC00ðxÞ � R0iðx; yÞ, the perturbation parameters

4B03ðdÞ of fuzzy reasoning algorithm 3 is defined as follows:

4B03ðdÞ ¼ inff½B0 � B00�jAij � ðdÞA0ij; A0j � ðdÞA00j ; Bi � ðdÞB0i; i ¼ 1; . . . ;m; j ¼ 1; . . . ; ng;

4DiðdÞ ¼ inff½Di � D0i�jAij � ðdÞA0ij; A0j � ðdÞA00j ; Bi � ðdÞB0i; i ¼ 1; . . . ;m; j ¼ 1; . . . ; ng:

Similarly, we could define 4B01ðdÞ, 4B0

2ðdÞ and 4B0

4ðdÞ.

Theorem 4.1. Let 4DiðdÞ ¼ ki ði ¼ 1; . . . ;mÞ. Then the following are valid:

(1) If \ ¼ ^, and ! is Lukasiewicz, Goguen, Mamdani or Kleene-Dienes implication, then

4B03ðdÞ ¼ 4B0

4ðdÞ ¼

Vmi¼1ki;

(2) Suppose \ is the product operator. If ! is Lukasiewicz, Goguen, Mamdani or Kleene-Dienes implication,

then 4B03ðdÞ ¼ 4B0

4ðdÞ ¼

Qmi¼1ki;

(3) Suppose \ is Lukasiewicz t-norm operator, and! is Lukasiewicz, Mamdani or Kleene-Dienes implication.

Then 4B03ðdÞ ¼ 4B0

4ðdÞ ¼ 0 _ ð

Pmi¼1ki � 1Þ.

Proof. Owing to the distributivity of continuous t-norm, Algorithm 3 is equivalent to Algorithm 4. And theabove results are obtained by Theorem 2.2. h

5. Example

We give an example to illustrate the application of the results of this paper.The load rotational speed of a kind of diesel motor is insufficient. There mainly have five malfunctions: the

valve spring breaks (symbolled as y1), the carbon deposit of oil spray nozzle stops up the hole(symbolled asy2), the oil pipe bursts (symbolled as y3Þ, the oil is spouted too late (symbolled as y4) and the injection pumpactuates the key to roll the key (symbolled as y5). And the portents are represented as follows: x1 (the heat hasexhausted), x2 (vibration), x3 (the torque anxiously falls), x4 (the flowing tubing head pressure of the machineis excessively low), x5 (the quantity of the oil machine consumption is big) and x6 (the rotational speed cannotcome up).

Based on fuzzy theory and combined with actual operation condition of diesel motor and expert’s experi-ence, the degree lyj

ðxiÞ is presented, which is on behalf of the degree of the portent xi corresponding to themalfunction yj, where, i ¼ 1; 2; . . . ; 6; j ¼ 1; 2; . . . ; 5. The fuzzy malfunction diagnostic matrix R is derivedas follows [15]:

lyjðxiÞ y1 y2 y3 y4 y5

x1 0.6 0.4 0.1 0.98 0.1x2 0.8 0.98 0.3 0.1 0.1x3 0.95 0.1 0.8 0.3 0.98x4 0.1 0.1 0.98 0.1 0.1x5 0.1 0.1 0.9 0.1 0.1x6 0.3 0.6 0.9 0.98 0.95


R ¼

0:6 0:4 0:1 0:98 0:1

0:8 0:98 0:3 0:1 0:1

0:95 0:1 0:8 0:3 0:98

0:1 0:1 0:98 0:1 0:1

0:1 0:1 0:9 0:1 0:1

0:3 0:6 0:9 0:98 0:95

0BBBBBBBBBB@

1CCCCCCCCCCA:

Now assume that the portent happens is represented by +1, and that the portent does not happen is rep-resented by +0. If the portents of diesel motor are as follows: x3 (the torque anxiously falls), x4 (the flowingtubing head pressure of the machine is excessively low) and x5 (the quantity of the oil machine consumption isbig), then the vector of the portents X is derived as follows: X ¼ ð0; 0; 1; 1; 1; 0Þ: Using fuzzy inference method,the reasoning vector X 0 of the malfunctions can be derived.

(1) If the fuzzy inference method is given by the max–min compositional inference, then we derive thatX 0 ¼ X � R ¼ ð0:95; 0:1; 0:98; 0:3; 0:98Þ. According to the arrangement principle that the malfunctionshappen from the big one to the small one, it assures that y3 and y5 happen possibly.Suppose X � ð0:9ÞX 1, R � ð0:9ÞR1 and ! is the Lukasiewicz implication. Then we could chooseX 1 ¼ ð0; 0; 0:9; 0:9; 0:9; 0Þ,

R1 ¼

0:6 0:4 0:1 0:98 0:1

0:8 0:88 0:3 0:1 0:1

0:85 0:2 0:9 0:3 0:88

0 0:2 0:88 0 0

0 0:2 0:8 0 0

0:2 0:6 0:9 0:98 0:85

0BBBBBBBBB@

1CCCCCCCCCA; R2 ¼

0:6 0:4 0:1 0:98 0:1

0:8 0:98 0:3 0:1 0:1

0:855 0:09 0:72 0:27 0:882

0:09 0:09 0:882 0:09 0:09

0:09 0:09 0:81 0:09 0:09

0:3 0:6 0:9 0:98 0:95

0BBBBBBBBB@

1CCCCCCCCCA:

Thus we derive the reasoning vector X 01 of the robust malfunctions is as follows: X 01 ¼ X 1 � R1 ¼ð0:85; 0:2; 0:9; 0:3; 0:88Þ. According to the arrangement principle that the malfunctions happen fromthe big one to the small one, it assures that y3 and y5 happen possibly. By Definition 1.1, we can calculatethat ½X 01 � X 0� ¼ 0:9.

(2) If the fuzzy inference methods is given by the max-product compositional inference, then we derive thatX 0 ¼ X � R ¼ ð0:95; 0:1; 0:98; 0:3; 0:98Þ. According to the arrangement principle that the phenomenonhappens from the big one to the small one, it assures that y3 and y5 happen possibly.

Suppose X � ð0:9ÞX 1, R � ð0:9ÞR1 and ! is the Goguen implication. Then we can choose X 1 ¼ð0; 0; 0:9; 0:9; 0:9; 0Þ, and R2 is given by the above. Thus we derive that the reasoning vector X 02 of the robustmalfunctions is as follows: X 02 ¼ X 2 � R2 ¼ ð0:7695; 0:081; 0:7938; 0:243; 0:7938Þ. According to the arrange-ment principle that the malfunctions happen from the big one to the small one, it assures that y3 and y5 happenpossibly. By the definition of 1.1, it can be calculated that ½X 01 � X 0� ¼ 0:81.


From the above discussion, it sees that the combination of min operation and Lukasiewicz implication israther robust, and the combination of product operation and Goguen implication is also robust. In fact,if the best perturbation parameters of fuzzy reasoning can be calculated, then the fuzzy reasoning is appli-cable.

6. Conclusions

Based on logically equivalence measure, we have defined the robustness of both fuzzy sets and fuzzy con-nectives in terms of d-equalities, which is more comprehensive than that based on the distance measure in caseof the unit interval [0,1]. From the above definition, we have discussed robustness of fuzzy reasoning andderived the best perturbation parameters of fuzzy reasoning. Some general and systematic results have beenderived and a related problem posed in [8] has been partially solved here. In comparison with the previouswork, our formulation of the problem is more systematic and practical. We transformed the problem ofrobustness of fuzzy reasoning to the task of robustness of the related fuzzy connectives and fuzzy implicationoperators. This equivalent problem is more suitable for a comprehensive analysis. The results reveal that therobustness of fuzzy reasoning is directly linked to the selection of fuzzy connectives and implication operators.

References

[1] K.Y. Cai, Robustness of fuzzy reasoning and d-equalities of fuzzy sets, IEEE Trans. Fuzzy Syst. 9 (5) (2001) 738–750.[2] Z. Cao, A. Kandel, Applicability of some fuzzy implication operators, Fuzzy Sets Syst. (31) (1989) 151–186.[3] O. Cordon, F. Herrera, A. Peregrln, Searching for basic properties obtaining robust implication operators in fuzzy control, Fuzzy Sets

Syst. (111) (2000) 231–251.[4] D. Dubois, J. Lang, H. Prade, Fuzzy sets in approximate reasoning, Parts 1 and 2, Fuzzy Sets Syst. 40 (1991) 143–244.[5] M. Dankova, On approximate reasoning with graded rules, Fuzzy Sets Syst. 158 (2007) 652–673.[6] P. Hajak, Metamathematics of Fuzzy Logic, Kluwer Academic Publisher, Dordrecht, 1998.[7] Y.M. Li, Analysis of Fuzzy System, Science Press, Beijing, 2005.[8] Y.M. Li, D.C. Li, W. Pedrycz, J.J. Wu, An approach to measure the robustness of fuzzy reasoning, Int. J. Intell. Syst. 20 (4) (2005)

393–413.[9] R.R. Yager, On some new classes of implication operators and their role in approximate reasoning, Inform. Sci. 167 (2004) 193–216.

[10] G.J. Wang, On the logic foundation of fuzzy reasoning, Inform. Sci. (117) (1999) 47–88.[11] M. Ying, Perturbation of fuzzy reasoning, IEEE Trans. Fuzzy Syst. 7 (5) (1999) 625–629.[12] L.A. Zadeh, The concept of a linguistic variable and its applications to approximate reasoning, I, II, III, Inform. Sci. 8 (1975) 199–

249, 301–357, 1975; 9(1975), 43–80.[13] L.A. Zadeh, A theory of approximate reasoning, in: J. Hayes, D. Michie, L.I. Mikulich (Eds.), Machine Intelligence, vol. 9, Halstead

Press, New York, 1979, pp. 149–194.[14] L.A. Zadeh, Toward a generalized theory of uncertainty(GTU)—an outline, Inform. Sci. 172 (2005) 1–40.[15] Z.Y. Zhou, The application of fuzzy technic in the malfunction diagnosis of diesel motor, Coal Mine Mach. 12 (2005) 162–164.

robustness of fuzzy reasoning via logically equivalence measure

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